# Are undecidable consequences of Con recursively enumerable?

Let $X\subset\Pi_1^0$ be the set of statements which are provable in PA$+$Con(PA) but independent of PA. Is $X$ recursively enumerable?

• Suggestion. Take also PA + not Con(PA). Then enumerate the theorems of both PA + Con(PA) and PA + non Con(PA). In the final enumeration, take the theorems that give opposite results in the two systems. Could that work? – Lucas K. May 2 '11 at 10:16
• What means "opposit results"? – Alex Gavrilov May 2 '11 at 10:45
• If you can prove p in PA + Con(PA) and not p in PA + not Con(PA), then p must be independent of PA (or PA is inconsistent). At that moment you can enumerate p in your solution. – Lucas K. May 2 '11 at 11:17
• Lucas, If I got you right, you suggest to enumerable the statements which are equivalent to Con. I suspect that there are statements which are strictly weaker then Con but still undecidable. – Alex Gavrilov May 2 '11 at 11:19
• Do you answer your own question now? – Lucas K. May 2 '11 at 12:32

The answer is no, and in particular, $X$ is $\Pi^0_1$-hard. Let $\sigma(x)=\exists v\,\theta(x,v)$ be a complete $\Sigma^0_1$-formula, where $\theta\in\Delta^0_0$, and find a formula $\pi(x)$ such that PA proves

$$\pi(x)\leftrightarrow\forall w\,(\mathrm{Proof_{PA}}(w,\ulcorner\pi(\dot x)\urcorner)\to\exists v\le w\,\theta(x,v))$$

by self-reference. Let $n\in\omega$. Since $\neg\pi(\bar n)$ is equivalent to a $\Sigma^0_1$ sentence, PA proves $\neg\pi(\bar n)\to\mathrm{Pr_{PA}}(\ulcorner\neg\pi(\bar n)\urcorner)$. By definition, $\neg\pi(\bar n)\to\mathrm{Pr_{PA}}(\ulcorner\pi(\bar n)\urcorner)$, hence PA proves $\mathrm{Con_{PA}}\to\pi(\bar n)$. I claim that

$$\tag{*}\mathbb N\models\sigma(n)\Leftrightarrow\mathrm{PA}\vdash\pi(\bar n),$$

which means that $n\mapsto\ulcorner\pi(\bar n)\urcorner$ is a reduction of the $\Pi^0_1$-complete set $\{n:\mathbb N\models\neg\sigma(n)\}$ to $X$.

To show $(*)$, assume first that $M\models\mathrm{PA}+\neg\pi(\bar n)$. Then there is no standard PA-proof of $\pi(\bar n)$, hence the witness $w\in M$ to the leading existential quantifier of $\neg\pi(\bar n)$ must be nonstandard. Then $\neg\theta(n,v)$ holds for all $v\le w$, and in particular, for all standard $v$, hence $\mathbb N\models\neg\sigma(\bar n)$.

On the other hand, assume that PA proves $\pi(\bar n)$, and let $k$ be the code of its proof. Since PA is sound, $\mathbb N\models\pi(\bar n)$, hence there exists $v\le k$ witnessing $\theta(\bar n,v)$, i.e., $\mathbb N\models\sigma(\bar n)$.

Here's a proof that doesn't directly diagonalize but instead relies on well-known results, which in turn were proved by diagonalization. So ultimately, it isn't really easier than Emil's, but it may be easier to find and remember.

I claim first that, if a $\Pi^0_1$ sentence $\phi$ is provable in PA plus $\neg\text{Con}(PA)$, then it is already provable in PA. This is probably well known, but here's a proof anyway. The assumption is equivalent to saying that $\text{Con}(PA)$ is provable from PA plus $\neg\phi$. But since $\neg\phi$ is a $\Sigma^0_1$ sentence, one can also prove from PA plus $\neg\phi$ that PA proves $\neg\phi$. Combining the preceding two sentences, we get a proof from PA plus $\neg\phi$ that PA plus $\neg\phi$ is consistent. By Gödel's second incompleteness theorem, it follows that PA plus $\neg\phi$ is inconsistent. This means that PA proves $\phi$, as claimed.

Now consider the transformation $T$ on $\Pi^0_1$ sentences defined by letting $T(\phi)$ be $\text{Con}(PA)\lor\phi$. I claim that $T(\phi)$ is in the set $X$ of the question if and only if PA does not prove $\phi$. To see this, note first that $T(\phi)$ is trivially provable from PA plus $\text{Con}(PA)$. So $T(\phi)\in X$ if and only if PA doesn't prove $\text{Con}(PA)\lor\phi$. That's if and only if PA plus $\neg\text{Con}(PA)$ doesn't prove $\phi$. And, by the claim proved above, that's if and only if PA doesn't prove $\phi$.

So T is a (trivially computable) many-one (in fact one-one) reduction to $X$ of the set of $\Pi^0_1$ sentences not provable in PA. The latter set is known not to be recursively enumerable; therefore neither is $X$.

• Thank you, Andreas. This is nice! Why didn't I see it before? – Alex Gavrilov May 3 '11 at 6:50
• Adnreas, how do we know that if $¬\varphi$ is $\Sigma_1^0$, then $PA+¬\varphi$ proves $Pr_{PA}(¬\varphi)$ ? – mtg May 24 '13 at 13:19
• PA proves $\sigma\to\Pr_{PA}(\ulcorner\sigma\urcorner)$ for every $\Sigma^0_1$-sentence $\sigma$ by formalizing the proof of $\Sigma^0_1$-completeness of Q. – Emil Jeřábek supports Monica May 24 '13 at 15:16
• It might be useful to add that this "provable $\Sigma^0_1$-completeness" result can be found in textbooks as an ingredient in the proof of the second incompleteness theorem. – Andreas Blass May 24 '13 at 16:33
• I didn't know how to make a comment, sorry, but you're right. In the meantime I figured out that $\Sigma^0_1$-completeness of $PA$ is provable in $PA$, but thanks - yeah, that's pretty obvious. – mtg May 24 '13 at 21:55